If frac{2 sin theta}{1+cos theta+sin theta}=y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $y = \frac{2 \sin 	heta}{1+\cos 	heta+\sin 	heta}$.Using half-angle formulas $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}

Vedclass · Accepted Answer

$y$

Vedclass · Answer

(A) Given $y = \frac{2 \sin 	heta}{1+\cos 	heta+\sin 	heta}$.Using half-angle formulas $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$,$\cos 	heta = 2 \cos^2 \frac{	heta}{2} - 1$,and $1 = \sin^2 \frac{	heta}{2} + \cos^2 \frac{	heta}{2}$:$y = \frac{2(2 \sin \frac{	heta}{2} \cos \frac{	heta}{2})}{2 \cos^2 \frac{	heta}{2} + 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}} = \frac{4 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{2 \cos \frac{	heta}{2}(\cos \frac{	heta}{2} + \sin \frac{	heta}{2})} = \frac{2 \sin \frac{	heta}{2}}{\cos \frac{	heta}{2} + \sin \frac{	heta}{2}}$.Now,consider the expression $E = \frac{1-\cos 	heta+\sin 	heta}{1+\sin 	heta}$.Using $1-\cos 	heta = 2 \sin^2 \frac{	heta}{2}$ and $1+\sin 	heta = (\sin \frac{	heta}{2} + \cos \frac{	heta}{2})^2$:$E = \frac{2 \sin^2 \frac{	heta}{2} + 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{(\sin \frac{	heta}{2} + \cos \frac{	heta}{2})^2} = \frac{2 \sin \frac{	heta}{2}(\sin \frac{	heta}{2} + \cos \frac{	heta}{2})}{(\sin \frac{	heta}{2} + \cos \frac{	heta}{2})^2} = \frac{2 \sin \frac{	heta}{2}}{\sin \frac{	heta}{2} + \cos \frac{	heta}{2}} = y$.

If $\frac{2 \sin \theta}{1+\cos \theta+\sin \theta}=y$,then $\frac{1-\cos \theta+\sin \theta}{1+\sin \theta}=$

Similar Questions

If $\sin 6\theta = 32\cos^5\theta \sin\theta - 32\cos^3\theta \sin\theta + 3x$,then $x = $

If $\frac{\cos ^2 48^{\circ}-\sin ^2 12^{\circ}}{\sin ^2 24^{\circ}-\sin ^2 6^{\circ}}=\frac{\alpha+\beta \sqrt{5}}{2}$, where $\alpha, \beta \in N$, then $\alpha+\beta$ is equal to . . . . . . .

If $\sin \alpha = \frac{-3}{5}$ where $\pi < \alpha < \frac{3\pi}{2},$ then $\cos \frac{\alpha}{2} = $

Prove that $\tan 4x = \frac{4 \tan x (1 - \tan^2 x)}{1 - 6 \tan^2 x + \tan^4 x}$.

If $\sin x + \cos x = \frac{1}{5},$ then $\tan 2x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module